# Why is work done on a ball lifting upward weight times distance and not greater?

Why is the work done on a ball lifting upward is its weight multiplied by distance moved ?

Shouldn't the force applied on the ball must be a value greater than its weight to make it move upward ? Why is it equal to the balls weight ?

• Work done does not equal energy expended to accomplish such work. Work just reflects the energy (in this case, stored) as a result of the ball's near position w.r.t. the gravity field it has moved in. Jan 15 '18 at 21:51
• so does it mean, in whatever the velocity and acceleration I moved the ball to a certain height. The work done(stored) is equal to weight x height ?
– Yogi
Jan 15 '18 at 21:54
• Yes. Some energy could be lost (and often is lost) through friction and other processes. Jan 15 '18 at 22:40
• Jan 15 '18 at 23:15

When you do work to equal out a force that is holding back, then "nothing" is holding back any more. You are free to move the object around by adding negligible amounts of extra energy on top of that, i.e. by doing no extra effort.

If you do more than just move the object to a higher place - if you e.g. give it kinetic energy as well - then you must indeed do more work than just to overcome gravity.

Think of gravity as "pulling energy out" of the system, so that you need to add exactly that extra amount to exactly balance the effect of gravity.

If the force is constant, and is greater than $mg$, the ball will be moving upward when it reaches the upper end of its path. The work then is indeed greater than the difference in potential energy: it is the sum of that and the kinetic energy the ball has by virtue of its motion.

On the other hand, if the force is initially greater than $mg$, in order to accelerate the ball upward, and later decreases to less than $mg$ so as to arrive at the upper location just as the speed reaches zero, the work is exactly $mgh$. In this case the work is the integral of the vertical force along the path.

The work you need to supply is independent of the path the object takes. Instead of directly lifting the object you could also, like in my diagram, first accelerate the mass along a straight path and then let it hit a ramp. I want to show that to lift the particle to some height $$h$$ you need to supply work equal to $$mgh$$ on the straight path. When the mass hits the ramp it will start moving upward and if the right velocity was given before it hit ramp it will exactly reach our desired height $$h$$.

How fast does the particle need to be moving to reach $$h$$? Let's call the speed of the mass before it hit ramp $$v_0$$, the speed when it's moving upwards $$v$$ and let's call the height of the mass $$y$$. The mass will decelerate due to gravity with $$a=g$$, so \begin{align}a&=g&(1)\\v(t)&=v_0-gt&(2)\\y(t)&=v_0t-\tfrac{1}{2}gt^ 2&(3)\end{align} The mass will reach the top at $$t=T$$, so using (2) that gives us $$T=\frac{v_0}{g}$$. Plugging that in in (3) and solving for $$v_0$$ gives $$v_0=\sqrt{2gh}$$ To reach this speed you need to supply work until its kinetic energy is equal to $$E=\tfrac{1}{2}mv_0^2=\tfrac{1}{2}m\cdot2gh=mgh$$ This was a long derivation and you don't need to understand all the steps exactly. What's important is that it doesn't matter in which way you supply work, it will always have the same result if no energy is lost.